For a statistical data x₁, x₂, …, x₁₀$x_1, x_2, \ldots, x_{10}$ of 10$10$ values, a student obtained the mean as 5.5$5.5$ and Σi=1¹⁰ xᵢ² = 371$\sum_{i=1}^{10} x_i^2 = 371$ . He later found that he had noted two values in the data incorrectly as 4$4$ and 5$5$, instead of the correct values 6$6$ and 8$8$, respectively. The variance of the corrected data is :

Related Formula

Core Logic

Given Mean = 8$= 8$ for 7 observations: 2, 4, 10, x, 12, 14, y$2, 4, 10, x, 12, 14, y$.

Given Variance = 16$= 16$:

Step 1: Solve for x and y

Using algebraic identity (x+y)² = x² + y² + 2xy$(x+y)^2 = x^2 + y^2 + 2xy$:

Since x+y=14$x+y=14$ and xy=48$xy=48$, roots of quadratic t² - 14t + 48 = 0$t^2 - 14t + 48 = 0$ are 8, 6$8, 6$. Given x > y$x > y$, we must select x = 8$x = 8$ and y = 6$y = 6$.

Step 2: Construct the set and evaluate Probability

The new set X$X$ is formed by 1, 2, 3, x-4, y, 5$\{1, 2, 3, x-4, y, 5\}$. Substituting x=8$x=8$ and y=6$y=6$, we get 1, 2, 3, 4, 6, 5$\{1, 2, 3, 4, 6, 5\}$. There are 6 distinct elements: 1, 2, 3, 4, 5, 6$\{1, 2, 3, 4, 5, 6\}$.

We choose two numbers without replacement. Total outcomes = 6 × 5 = 30$6 \times 5 = 30$ permutations (or 62 = 15$\binom{6}{2} = 15$ combinations). Let's use combinations. Total ways to choose 2 numbers = 62 = 15$\binom{6}{2} = 15$.

We need the probability that the smaller number is less than 4$4$. P(smaller < 4) = 1 - P(smaller ≥ 4)$P(\text{smaller} < 4) = 1 - P(\text{smaller} \geq 4)$.

Step 3: Final Calculation via Complement

For the smaller number to be ≥ 4$\geq 4$, BOTH chosen numbers must be ≥ 4$\geq 4$. The available numbers ≥ 4$\geq 4$ in the set are 4, 5, 6$\{4, 5, 6\}$ (Total 3 numbers). Ways to choose two numbers from these 3 is 32 = 3$\binom{3}{2} = 3$.

Pattern Recognition

When statistical problems ask for "at least one" or "minimum bounding", calculating the complement probability (e.g., both elements strictly exceeding the threshold) cuts combinatorial checks from 3+$3+$ cases down to exactly 1$1$.

Chapter Mix

Class 11 Maths: Statistics Class 12 Maths: Probability

Related Formula

Mean deviation about median is given by:

Core Logic

For 1000$1000$ numbers in A.P., median XM = (1001k)/(2)$X_M = \frac{1001k}{2}$.

Step 1: Evaluation of k and k^2

Given mean deviation is 500$500$:

Therefore, k² = 4$k^2 = 4$.

Pattern Recognition

Use symmetry of A.P. around median to quickly evaluate absolute deviation sum.

Chapter Mix

Class 11 Maths: Statistics

Related Formula

Core Logic

Calculate x+y$x+y$ using the mean formula:

Step 1: Calculate sum of squares using Variance

Using the variance formula:

Step 2: Solve for x and y

From (1), y = 19 - x$y = 19 - x$. Substitute into (2):

Thus, x = 12$x = 12$ and y = 7$y = 7$ (or vice-versa, which doesn't affect absolute differences).

Step 3: Calculate the New Mean

We need the mean of 4$4$ numbers: x, y, x + y + 1, |x - y|$x, y, x + y + 1, |x - y|$. Substitute x = 12, y = 7$x = 12, y = 7$: The numbers are 12, 7, (12+7+1), |12-7| = 12, 7, 20, 5$12, 7, (12+7+1), |12-7| = 12, 7, 20, 5$. New Mean = (12 + 7 + 20 + 5)/(4) = (44)/(4) = 11$\frac{12 + 7 + 20 + 5}{4} = \frac{44}{4} = 11$.

Pattern Recognition

Statistics questions mapping x+y$x+y$ and x²+y²$x^2+y^2$ inherently hide a quadratic symmetric system (x+y)² - 2xy = x²+y²$(x+y)^2 - 2xy = x^2+y^2$. Solving for the roots immediately grants the discrete terms.

Chapter Mix

Class 11 Maths: Statistics

Related Formula

Core Logic

Midpoints (xᵢ$x_i$): 6, 10, 14, 18$6, 10, 14, 18$. Frequencies (fᵢ$f_i$): 3, λ, 4, 7$3, \lambda, 4, 7$. Sum of frequencies Σ fᵢ = 3 + λ + 4 + 7 = 14 + λ$\sum f_i = 3 + \lambda + 4 + 7 = 14 + \lambda$.

Calculate Mean (μ$\mu$):

Since data is standard, μ$\mu$ and λ$\lambda$ typically hold integer values, hinting λ + 14$\lambda + 14$ must be a factor of 60. Therefore λ$\lambda$ could be 1, 6, 16$1, 6, 16$.

Step 1: Applying Variance

Variance σ² = 19$\sigma^2 = 19$.

Substitute into variance:

Step 2: Testing Candidates

Let's test the potential integer λ = 6$\lambda = 6$ (a factor of 60 is 14+6=20$14+6=20$): If λ = 6$\lambda = 6$:

Let's verify variance for λ = 6$\lambda = 6$:

This perfectly matches the given variance.

Thus, λ = 6$\lambda = 6$ and μ = 13$\mu = 13$.

Pattern Recognition

Instead of solving a horrible cubic/quadratic equation for the variance parameter, use the integer division trait of the mean to test valid candidates.

Chapter Mix

Class 11 Maths: Statistics

Related Formula

Core Logic

Let the 9 unchanged numbers be x₁, x₂, , x₉$x_1, x_2, \dots, x_9$. Case 1 (with α$\alpha$):

Step 1: Applying Replacement Conditions

Case 2 (with β$\beta$): Mean = 10.1

Variance = 1.99

Step 2: Solving for Alpha and Beta

Substitute Σi=1⁹ xᵢ² = 1020 - α²$\sum_{i=1}^9 x_i^2 = 1020 - \alpha^2$:

Since β - α = 1$\beta - \alpha = 1$:

(Solving gives α = 19/2, β = 21/2$\alpha = 19/2, \beta = 21/2$).

Pattern Recognition

When a single term is replaced, immediately establish the difference of sums and difference of sum-of-squares. The identity β² - α²$\beta^2 - \alpha^2$ gracefully factors to use the established β - α$\beta - \alpha$ value.

Chapter Mix

Class 11 Maths: Statistics